# How do you solve (8x)^(1/2)+6=0?

Jul 27, 2016

$x = \frac{9}{2}$

$x = 4.5$

#### Explanation:

${\left(8 x\right)}^{\frac{1}{2}} + 6 = 0$

Get rid of 6 from left side
For that subtract 6 on both sides

${\left(8 x\right)}^{\frac{1}{2}} = - 6$

Squaring on both sides

$8 x = 36$

$x = \frac{36}{8}$

$x = \frac{9}{2}$

$x = 4.5$

Jul 28, 2016

There are no values of $x$ which satisfy this equation.

#### Explanation:

${\left(8 x\right)}^{\frac{1}{2}} + 6 = 0$

Subtract $6$ from both sides to get:

${\left(8 x\right)}^{\frac{1}{2}} = - 6$

Square both sides, noting that squaring may introduce spurious solutions:

$8 x = 36$

Divide both sides by $8$ to get:

$x = \frac{36}{8} = \frac{9}{2}$

Check:

${\left(8 x\right)}^{\frac{1}{2}} + 6 = {\left(8 \cdot \frac{9}{2}\right)}^{\frac{1}{2}} + 6 = {36}^{\frac{1}{2}} + 6 = 6 + 6 = 12$

So this $x$ is not a solution of the original equation.

The problem is that while $36$ has two square roots (viz $\pm 6$), ${36}^{\frac{1}{2}} = \sqrt{36} = 6$ denotes the principal, positive square root.

So the original equation has no solutions (Real or Complex).